Can you find area of the Green shaded Triangle? | (Two Methods) | #math #maths | #geometry

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Can you find area of the Green shaded Triangle? | (Two Methods) | #math #maths | #geometry

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[davidellis]

Here's a different kind of solution. The locus of the right angle of the yellow triangle is a semicircle inside the square whose diameter is the left side of the square. If we assume that the area of the green triangle is determined uniquely by the available data, then we can safely place the right angle at the center of the square, making the green triangle an isosceles right triangle with legs 2 and area (1/2)(2)(2), which is 2. The solutions given in the video confirm the above assumption.

[redfinance]

I like this problem very much! Here’s my solution : let x be the side length of the square. draw a perpendicular to AD from E, let the intersection at AD be labelled as G. Let GE = h. draw another perpendicular to AB from E and label its intersection at AB as H. HE = sqrt(2^2 - h^2) = sqrt(4 - h^2) by Pythagoras. The area of triangle AED is either (AD)(GE)/2 or (AE)(ED)/2 so (AD)(GE)/2 = (AE)(ED)/2, so ED = xh/2. By Pythagoras, ED^2 + AE^2 = AD^2. So (xh/2)^2 + 2^2 = x^2. Solving for h^2 gives h^2 = 4 - 16/x^2. Now, HE = sqrt(4 - h^2) = sqrt(16/x^2) = 4/x. Finally, area of green triangle equal to (AB)(HE)/2 = 1/2 * x * 4/x = 2 units squared! When you can’t solve with similar triangles, look for Pythagoras. When you can’t solve by Pythagoras, solve by trigonometry. When you can’t solve by trigonometry, solve by coordinate geometry 😂

[yatesfletcher]

If F is the projection of E onto AB, let y be the length of EF. The area G of the green triangle is 1/2*x*y. Triangle AEF is similar to triangle DAE. Thus 2/x=y/2 => y=4/x so G=1/2*x*(4/x)=2.

[montynorth]

DE can also be 2 units which fully satisfies the 90 degree angle requirement, and then each yellow corner angle will be 45 degrees.
Yellow and green will be congruent with areas 1/2 x 2 x 2.
2 units.

[timmcguire]

Kinda solved in my head in 10 seconds: Let E be the center point of the square making the diagonals 4, the sides 2sqrt2, and each 1/4 of the square, one of which is our triangle, 2 square units..

[phungpham]

1/Drop the height BH to AE (extended)
We have: angle AEE = angle BAH ( having sides perpendicular) so the two triangles ADE and HAB are congruent so, AE=BH=2
—-> Area of the green triangle= 1/2 x2x2= 2 sq units😅

[marcgriselhubert]

Let's use an orthonormal center A and first axis (AD)
We have A(0; 0) D(c; 0) and E(2.cos(t); 2.sin(t))
with c the side length of the square and t = angleEAD
Then VectorAE(2.cos(t); 2.sin(t))
and Vector(ED(2.cos(t) -c; 2.sin(t)
These vectors are orthogonal, so we have:
(2.cos(t)).(2.cos(t) -c) +(2.sin(t)).(2.sin(t)) = 0,
or 4 -2.c.cos(t) = 0 or c.cos(t) = 2
The basis of the green triangle is AB = c, the corresponding height is the abscissa of E: 2.cos(t), so its area is
(1/2).(c).(2.cos(t)= = c.cos(t) = 2

[tellerhwang]

1.extend AE→AH丄BH
△ADE, △BAH congruent(AAS)
→AE=BH=2
2.△AEB=AE·BH/2=2😊

[murdock]

Nice! ∎ABCD → AB = AF + BF = BC = CD = AM + DM = r + r; AE = 2; sin⁡(EFA) = 1;
EF = h; ADE = FAE = δ → sin⁡(δ) = 2/2r = 1/r = h/2 → hr = 2 → h = 2/r → rh = 2 = area ∆ AEB

[jimlocke]

DE is not given so the solution must be the same for all valid values of DE and we can solve special cases. monthnorth3009 chose DE = 2 and found the area of the green triangle to be 2. A simpler special case is DE = 0. ΔADE is now a line segment equal to the side of the square, so AD = AB = 2 and green triangle has area 2.

However, we no longer really have a triangle. To make our case more legitimate, let DE be extremely small but >0. As we make DE smaller, we can make AD as close as we want to AE but never make them equal as long as DE > 0. Eventually, we can make DE small enough to treat the difference between AD and AE as negligible. In mathematics, AD = AE is called the limit as DE approaches 0.

[sergeyvinns]

Так это тот же болт. только вид с боку! h/2 = 2/x=sina, тоже тригонометрия.

[sergeyvinns]

Интересно. какой второй метод. Первый решается устно!

[LuisdeBritoCamacho]

I am going to be very quick

01) Draw a Vertical Line from E. Define Point F between Point A and Point B.
02) Triangle ADE is congruent with Triangle AEF
03) 2 / AD = EF / 2
04) AD * EF = 4
05) AD = AB
06) Green Area = 4 / 2 = 2

ANSWER : Area of Green Triangle equal to 2 Square Units.

[scottwebster]

Don't put any advanced math, physics, engineering degrees on job applications. Sometimes you just need a job to pay the bills and they won't hire you as a truck driver if you are being honest..

[misterenter-izrz]

Usual trick, we suppose the diagonal is 4, side is 4/sqrt(2), area is 1/4×(4/sqrt(2))^2=2.😂😂😂